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Speeding and police car

  1. Aug 31, 2011 #1
    A speeder passes a parked police car at 30.0 m/s. The police car starts from rest with a uniform acceleration of 2.60 m/s2.

    (a) How much time passes before the police car reaches the speeder?

    (b) How far does the speeder get before being overtaken by the police car?




    Relevant equations:
    v=vo+at

    v2=vo2+2a[itex]\Delta[/itex]x


    well...i drew a picture of the situation and wrote the givens...but i'm not sure where to go from there...

    i mainly want HOW to solve this problem rather than the answer itself...
     
  2. jcsd
  3. Aug 31, 2011 #2

    Doc Al

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    Staff: Mentor

    Write the position as a function of time for each. (You'll need a different kinematic formula.)
     
  4. Aug 31, 2011 #3
    you need to equal the distances of the police car and the car.

    What i mean is, solve for distance with both and join the equations (as both equal distances), then solve for time...
     
  5. Aug 31, 2011 #4
    so i should use x=vot+0.5at^2?
     
  6. Aug 31, 2011 #5

    Doc Al

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    Staff: Mentor

    Yes.
     
  7. Aug 31, 2011 #6
    p=police
    s=speeder

    so would i do this?

    vopt+[itex]\frac{1}{2}[/itex]apt=vost+[itex]\frac{1}{2}[/itex]ast

    because the time will be the same, right?

    and then i use the "t" time and find the displacement?
     
  8. Aug 31, 2011 #7
    i meant .5apt^2 and .5ast^2
     
  9. Aug 31, 2011 #8

    Doc Al

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    Good. So what is the initial velocity and the acceleration for each?
     
  10. Aug 31, 2011 #9
    Here’s what Galileo stated (verbatim) in the Chapter titled THIRD DAY, which pertains to your scenario:

    [208]

    THEOREM I, PROPOSITION I

    “The time in which any space is traversed by a body starting from rest and uniformly accelerated is equal to the time in which that same space would be traversed by the same body moving at a uniform speed whose value is the mean of the highest speed and the speed just before acceleration began.”

    By manner of the associated proportionalities of Galileo’s Theorem I, Proposition I, it is realized that you’ve been given the uniform speed (constant velocity) by which Galileo refers to as the mean of the highest and lowest velocities that will be achieved by the accelerating police car (0 m/s and 60 m/s) per this scenario. Since 30 m/s is the given mean velocity of the car at its constant velocity, it’s realized that 60 m/s is the accelerating police car’s highest velocity.

    Therefore, the quickest way to derive the intercept distance (s) is by applying the 60 m/s velocity (v) that will be achieved by the police car per its rate of acceleration (a) via the following simplistic kinematics equation:

    s = (v)^2 / 2 / a

    If you afterward apply the appropriate kinematics equation per use of the intercept distance (s) and acceleration (a), you can easily derive the intercept time (t).

    You will find that the time (t) of the accelerating police car and the time (t) required by the car traveling at constant velocity are the same time (t) just as both vehicles have traversed the same distances (s) upon intercept hence, the nature of Galileo’s Theorem I, Proposition I, and the reason I provided it.

    If you had applied the given 30 m/s constant velocity (v) of the car to derive the distance (s) at which the police car would finally match the velocity of the car (not intercept), it is a simple matter of quadrupling that distance (s) to derive the intercept distance (s).
     
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